Subtract 14 from the product of 3 and n
1 answer:
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Answer:
- 0.5 + 2.985i
- 1 + 2.828i
- 1.5 + 2.598i
- 2 + 2.236i
Explanation:
Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.
Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.
The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:
Then, the work consists in finding pairs (a,b) for which:
You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:
I will use b =0.5, b = 1, b = 1.5, b = 2
Then, four distinct complex numbers that have an absolute value of 3 are:
- 0.5 + 2.985i
- 1 + 2.828i
- 1.5 + 2.598i
- 2 + 2.236i
Answer:
plug in the numbers for radius, and know your formulas. a diameter is 2 times a radius so
Pi or 3.14 times D is the same as 2 times pi times radius.
